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Critical Thinking Consider the following Minitab display

Understandable Statistics | 9th Edition | ISBN: 9780618949922 | Authors: Charles Henry Brase, Corrinne Pellillo Brase ISBN: 9780618949922 213

Solution for problem 3 Chapter 3

Understandable Statistics | 9th Edition

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Understandable Statistics | 9th Edition | ISBN: 9780618949922 | Authors: Charles Henry Brase, Corrinne Pellillo Brase

Understandable Statistics | 9th Edition

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Problem 3

Critical Thinking Consider the following Minitab display of two data sets. Variable N Mean SE Mean StDev Minimum Q1 Median Q3 Maximum C1 20 20.00 1.62 7.26 7.00 15.00 20.00 25.00 31.00 C2 20 20.00 1.30 5.79 7.00 20.00 22.00 22.00 31.00 (a) What are the respective means? the respective ranges? (b) Which data set seems more symmetric? Why? (c) Compare the interquartile ranges of the two sets. How do the middle halves of the data sets compare?

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Molecular Week 8 Notes  Methods to determine DNA sequence recognized by a gene regulating protein o Mixed with short fragments of DNA o Double stranded DNA fragments that bind to regulating proteins are separated through gel mobility shifts o Separation of DNA-protein complexes from free DNA o DNA fragments removed from protein  more rounds o Nucleotide sequence can be determined and a consensus DNA recognition sequence can be made  DNA footprinting o Protein bound DNA radioactively labeled o DNA cleaved (random single stranded cuts) o Protein removed o DNA denatured and separated into two strands o Resultant fragments from labeled strand  Alignments o Look for highly conserved regions in a promoter (where transcription factors bind)  Chip assay to identify transcription factors o Chromatin immunoprecipitation o Crosslink between gene regulating protein and DNA  chop DNA  isolate fragments with antibody against the protein (precipitate)  reverse crosslinks and remove protein left with DNA sequence that protein binds to (amplify) o Allows identification of all the sites in a genome that a regulatory protein occupies o The identities for the precipitated, amplified DNA fragments can be determined by hybridizing the mixture of fragments to DNA microarrays  Gene regulatory circuit o The complete set of genes controlled by 3 key regulatory proteins in budding yeast, as deduced from the DNA sites where the regulatory proteins bind o Reg. proteins- Mata 1, Matα1, and Matα2  Specify the two different haploid mating types (analogous to male and female gametes) of this unicellular organism o 16 chromosomes in the yeast genome- show where various combinations of the 3 regulatory proteins bind o Proteins act in complexes to regulate genes o Determinations of complete transcriptional circuits show that transcriptional networks are not infinitely complex o Chip analysis and phylogenetic footprinting  Operons- prokaryotes o Operon- transcriptional unit where all of the genes responsible for 1 process are together regulated by 1 promoter o TATA box and consensus sequence- sigma factor binds with polymerase o Operator- binds factor that gets in the way and prevents sigma factor and polymerase from binding which turns off the genes o Clustered genes in E. coli that code for enzymes that manufacture tryptophan  5 genes of the Trp operon are transcribed as a single mRNA molecule which allows their expression to be controlled coordinately  Clusters of genes transcribed as a single mRNA molecule are common in bacteria; each cluster is called an operon  Activators and repressors o If the level of tryptophan inside the cell is low, RNA polymerase binds to the promoter and transcribes the 5 genes of the Trp operon o If the level of tryptophan is high, the trp repressor is activated to bind to the operator where it blocks the binding of RNa polymerase to the promoter o Whenever the level of intracellular Trp drops, the repressor releases its tryptophan and becomes inactive which allows the polymerase to begin transcribing the genes  Tryptophan repressor o The binding of Trp to the tryptophan repressor protein changes its conformation o This structural change enables the gene regulatory protein to bind tightly to a specific DNA sequence (the operator), which blocks transcription of the genes encoding the enzymes required to produce tryptophan (the Trp operon) o Helix turn helix protein structure o Tryptophan binding increases the distance between the two recognition helices in the homodimer, allowing the repressor to fit snugly on the operator o Noncovalent bonds b/w trp and repressor o Trp binds  conformational change  recognition helix moves and can bind to major groove  Mechanism of gene regulatory proteins controlling gene transcription (prokaryotes) o Negative regulation o Positive regulation o Addition of a ligand can turn on a gene either by removing a gene repressor protein from the DNA or by causing a gene activator protein to bind o The addition of an inhibitory ligand can turn off a gene either by removing a gene activator protein from the DNA or by causing a gene repressor protein to bind o Activator protein helps attract RNA polymerase  Lambda repressor acting as an activator or repressor o Some bacterial gene regulatory proteins can act as either a transcriptional activator or a repressor, depending on the precise placement of their DNA-binding sites o Operator location- need correct distance from other sequence to let RNA polymerase bind o Ligand independent o Transcription is activated or repressed by lambda repressor (depends on conformation)  Dual control of the Lac operon o No cAMP then CAP can’t bind  Repressor not bound  No lactose o Glucose present  Prevents cAMP and CAP binding  Repressor binds in absence of ligand o No lactose  repressor bound No glucose  increase cAMP and CAP o No glucose  no repressor Lactose  CAP bound, increase cAMP o Glucose and lactose levels control the initiation of transcription of the Lac operon through their effects on CAP and the Lac repressor protein o LacZ, the first gene of the Lac operon, encodes the enzyme beta- galactosidase, which breaks down the disaccharide lactose to galactose and glucose o Lactose addition increases the concentration of allolactose, an isomer of lactose, which binds to the repressor protein and removes it from the DNA o Glucose addition decreases the concentration of cyclic AMP  Because cyclic AMP no longer binds to CAP, this gene activator protein dissociates from the DNA, turning off the operon o Several Lac repressor binding sites located at different positions along the DNA o The expression of the Lac operon never completely shuts down  A small amount of the enzyme beta-galactosidase is required to convert lactose to allolactose, thereby permitting the Lac repressor to be inactivated when lactose is added to the growth medium  DNA looping can stabilize protein-DNA interactions o Lac repressor is a tetramer- has multiple binding sites- can bind to 2 operators simultaneously  enhances repressor ability o The Lac operon has a total of three operators o At the concentrations of Lac repressor in the cell, and in the absence of lactose, the state in which 2 operators are bound is the most stable  To dissociate completely from the DNA, the Lac repressor must first pass through an intermediate where it is bound to only a single operator  The local concentration of the repressor is then very high in relation to the free operator, and the reaction to the double- bound form is favored over the dissociation reaction  Then even a low-affinity site can increase the occupancy of a high-affinity site and give higher levels of gene repression in the cell  Binding of two proteins to separate sites on the DNA double helix can greatly increase their probability of interaction o There is an optimal distance for DNA looping (about 500 bp) o Increases frequency of collision  Gene activation at a distance o Enhancers present hundreds of bp away- can loop around and encourage transcription o NtrC is a bacterial gene regulatory protein that activates transcription by directly contacting RNA polymerase and causing a transition between the initial DNA-bound form of the polymerase and the transcriptionally competent form o The transition stimulated by NtrC requires energy from ATP hydrolysis o Interaction of NtrC and RNA polymerase with the intervening DNA looped out  Interchangeable RNA polymerase subunits as a strategy to control gene expression in a bacterial virus o SPO1 uses a bacterial polymerase to transcribe its early genes immediately after the viral DNA enters the cell o 28 early gene encodes a sigmalike factor that binds to RNA polymerase and displaces the bacterial sigma factor  Initiates transcription of the SPO1 middle genes o Middle gene 34 displaces the 28 and directs RNA polymerase to transcribe the late genes  Produces the proteins that package the virus chromosome into a virus coat and lyse the cell  Sets of virus genes are expressed in the order in which they are needed  rapid and efficient viral replication  Gene control region- eukaryotes o Gene transcription factors and other regulating proteins can be adjacent to promoter or at a great distance upstream (can also be downstream in introns) o Hetero and dimeric proteins o The promoter- DNA sequence where TF and polymerase assemble o Regulatory sequences serve as binding sites for gene regulatory proteins, whose presence on the DNA affects the rate of transcription initiation o DNA looping allows gene regulatory proteins bound at any of these positions to interact with the proteins that assemble at the promoter o Many gene regulatory proteins act through a mediator while others influence the general transcription factors and RNA polymerase directly o Many gene regulatory proteins also influence the chromatin structure of the DNA control region which affects transcription initiation indirectly o The mediator and general transcription factors are the same for all polymerase II transcribed genes but the gene regulatory proteins and the locations of their binding sites relative to the promoter differ for each gene  The modular structure of a gene activator protein o Independent DNA-binding and transcription-activating domains in the yeast gene activator protein Gal4 o A functional activator can be reconstituted from the C-terminal portion of the yeast Gal4 protein if it is attached to the DNA-binding domain of a bacterial gene regulatory protein (LexA) by genetic engineering techniques o Resulting bacterial-yeast hybrid produced in yeast cells  activates transcription from yeast genes provided that the specific DNA-binding site for the bacterial protein has been inserted next to them o Gal4- responsible for activating the transcription of yeast genes that code for the enzymes that convert galactose to glucose o Chimeric gene regulatory protein requires a LexA recognition sequence to activate transcription o The control gene for a gene controlled by LexA was fused to the E. coli LacZ gene which codes for the enzyme beta-galactosidase (monitor expression level specified by a gene control region) o LacZ is a reporter gene- reports the activity of a gene control region o Phosphorylated galactose  activate gene with Gal4 TF regulating protein  2 domains- binds DNA consensus sequence and transcription activator domain o LacZ- gene construct under control of promoter that initially contained Gal 4  took off Gal4 DNA binding domain and put on LexA o Correct TAD but incorrect binding domain  no transcription  Fits into major groove but can’t H bond or interact o Need TAD and DNA binding domain for transcription  4 ways eukaryotic activator proteins can direct local alterations in chromatin structure to stimulate transcription initiation o Compact DNA- genes are tucked away o Activator protein with access to DNA close to gene  Can bind chromatin remodeling complexes, unravel, remove histones, or rearrange  expose gene  modify histone tails  unravel o Acetylation of histones makes it easier for histone chaperones to remove them from nucleosomes o Nucleosome remodeling and histone removal favor transcription initiation by increasing the accessibility of DNA and thereby facilitating the binding of Mediator, RNA polymerase, and the general TF and activator proteins o Transcription initiation and the formation of a compact chromatin structure can be regarded as competing biochemical assembly reactions and enzymes that increase the accessibility of DNA in chromatin will tend to favor transcription initiation  Writing and reading histone code during transcription initiation o Gene regulating protein binds to exposed DNA o Specific pattern triggered by 1 activator protein  3 modifications  exposes gene o Gene activator protein binds to DNA packaged into chromatin  attracts a histone acetyl transferase to acetylate lysine  histone kinase phosphorylates a serine  signals histone acetyl transferase rxn  histone code for transcription initiation is written o Writing is sequential- each histone modification depends on a prior modification o Final reading- general TF TFIID and chromatin remodeling complex bind  recognize acetylated histone tails through a bromodomain  Transcriptional synergy- activator proteins o Synergy- additive effect of multiple activators; multiple proteins cause huge increases in transcription o Observed b/w different gene activator proteins from the same organism and b/w activator proteins from different eukaryotic species when they are experimentally introduced into the same cell o High degree of conservation responsible for eukaryotic transcription initiation  6 ways in which eukaryotic gene repressor proteins can operate o Get in the way  Whichever substrate gets there first is expressed at a higher level  Competitive DNA binding o Eliminate activity of TAD  Both activator and repressor bind to DNA then bind to each other  Activator can’t do its job and bind to other proteins (chromatin remodeling complex/histone modification complex)  Masking the activation surface o Interfere with general TF  Both activator and repressor bind  Repressor prevents TF from coming together  Direct interaction with the general TF o Bind chromatin remodeling complex  Becomes more densely packed  not accessible  Growth arrest o Modify histone tails  Keep chromatin dense  Histone deacetylases o Modifications package and hide genes  Histone methyl transferase  histone methylation  proteins bind to methylated histones and hide genes  Coactivators/repressors o Bind to proteins that bind DNA o Given TF can fall into either category (on/off) depending on context it is used

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Chapter 3, Problem 3 is Solved
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Textbook: Understandable Statistics
Edition: 9
Author: Charles Henry Brase, Corrinne Pellillo Brase
ISBN: 9780618949922

This full solution covers the following key subjects: . This expansive textbook survival guide covers 57 chapters, and 994 solutions. This textbook survival guide was created for the textbook: Understandable Statistics, edition: 9. Since the solution to 3 from 3 chapter was answered, more than 559 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 3 from chapter: 3 was answered by , our top Statistics solution expert on 01/04/18, 09:09PM. The answer to “Critical Thinking Consider the following Minitab display of two data sets. Variable N Mean SE Mean StDev Minimum Q1 Median Q3 Maximum C1 20 20.00 1.62 7.26 7.00 15.00 20.00 25.00 31.00 C2 20 20.00 1.30 5.79 7.00 20.00 22.00 22.00 31.00 (a) What are the respective means? the respective ranges? (b) Which data set seems more symmetric? Why? (c) Compare the interquartile ranges of the two sets. How do the middle halves of the data sets compare?” is broken down into a number of easy to follow steps, and 78 words. Understandable Statistics was written by and is associated to the ISBN: 9780618949922.

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Critical Thinking Consider the following Minitab display